Problem: The equation of hyperbola $H$ is $\dfrac {(x+2)^{2}}{9}-\dfrac{y^2}{64} = 1$. What are the asymptotes?
Explanation: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac{y^2}{64} = - 1 + \dfrac {(x+2)^{2}}{9}$ Multiply both sides of the equation by $64$ $y^2 = { - 64 + \dfrac{ (x+2)^{2} \cdot 64 }{9}}$ Take the square root of both sides. $\sqrt{y^2} = \pm \sqrt { - 64 + \dfrac{ (x+2)^{2} \cdot 64 }{9}}$ $ y = \pm \sqrt { - 64 + \dfrac{ (x+2)^{2} \cdot 64 }{9}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y \approx \pm \sqrt {\dfrac{ (x+2)^{2} \cdot 64 }{9}}$ $y \approx \pm \left(\dfrac{8 \cdot (x + 2)}{3}\right)$ Rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{8}{3}(x + 2)+ 0$